| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard C4 vectors question testing routine techniques: finding a line equation from two points, showing lines intersect by solving simultaneous equations, and using perpendicularity conditions. All parts follow textbook methods with no novel insight required. Part (d) requires slightly more steps but is still algorithmic. Slightly easier than average due to straightforward application of standard procedures. |
| Spec | 1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\overrightarrow{AB} = (8\mathbf{i} - 6\mathbf{k}) - (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) = (-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\) | M1 | |
| \(\therefore \mathbf{r} = (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) + \lambda(-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\) | A1 | |
| (b) \(3 - 3\lambda = -2 + 7\mu\) ... (1) | ||
| \(6 + 2\lambda = 10 - 4\mu\) ... (2) | ||
| \(-8 + 2\lambda = 6 + 6\mu\) ... (3) | B1 | |
| (3) - (2): \(-14 = -4 + 10\mu, \quad \mu = -1, \quad \lambda = 4\) | M1 A1 | |
| check (1): \(3 - 12 = -2 - 7\), true \(\therefore\) intersect | B1 | |
| (c) \(\mathbf{r} = (-2\mathbf{i} + 10\mathbf{j} + 6\mathbf{k}) - (7\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}) \quad \therefore (-9, 14, 0)\) | M1 A1 | |
| (d) \(\overrightarrow{OC} = [(-2 + 7\mu)\mathbf{i} + (10 - 4\mu)\mathbf{j} + (6 + 6\mu)\mathbf{k}]\) | ||
| \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(-5 + 7\mu)\mathbf{i} + (4 - 4\mu)\mathbf{j} + (14 + 6\mu)\mathbf{k}]\) | M1 A1 | |
| \(\therefore [(-5 + 7\mu)\mathbf{i} + (4 - 4\mu)\mathbf{j} + (14 + 6\mu)\mathbf{k}] \cdot (-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 0\) | M1 | |
| \(15 - 21\mu + 8 - 8\mu + 28 + 12\mu = 0\) | ||
| \(15 - 21\mu + 8 - 8\mu + 28 + 12\mu = 0\) | A1 | |
| \(\mu = 3 \quad \therefore \overrightarrow{OC} = (19\mathbf{i} - 2\mathbf{j} + 24\mathbf{k})\) | M1 A1 | (14) |
**(a)** $\overrightarrow{AB} = (8\mathbf{i} - 6\mathbf{k}) - (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) = (-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ | M1 |
$\therefore \mathbf{r} = (3\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}) + \lambda(-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ | A1 |
**(b)** $3 - 3\lambda = -2 + 7\mu$ ... (1) |
$6 + 2\lambda = 10 - 4\mu$ ... (2) |
$-8 + 2\lambda = 6 + 6\mu$ ... (3) | B1 |
(3) - (2): $-14 = -4 + 10\mu, \quad \mu = -1, \quad \lambda = 4$ | M1 A1 |
check (1): $3 - 12 = -2 - 7$, true $\therefore$ intersect | B1 |
**(c)** $\mathbf{r} = (-2\mathbf{i} + 10\mathbf{j} + 6\mathbf{k}) - (7\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}) \quad \therefore (-9, 14, 0)$ | M1 A1 |
**(d)** $\overrightarrow{OC} = [(-2 + 7\mu)\mathbf{i} + (10 - 4\mu)\mathbf{j} + (6 + 6\mu)\mathbf{k}]$ |
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(-5 + 7\mu)\mathbf{i} + (4 - 4\mu)\mathbf{j} + (14 + 6\mu)\mathbf{k}]$ | M1 A1 |
$\therefore [(-5 + 7\mu)\mathbf{i} + (4 - 4\mu)\mathbf{j} + (14 + 6\mu)\mathbf{k}] \cdot (-3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 0$ | M1 |
$15 - 21\mu + 8 - 8\mu + 28 + 12\mu = 0$ |
$15 - 21\mu + 8 - 8\mu + 28 + 12\mu = 0$ | A1 |
$\mu = 3 \quad \therefore \overrightarrow{OC} = (19\mathbf{i} - 2\mathbf{j} + 24\mathbf{k})$ | M1 A1 | (14) |The line $l_1$ passes through the points $A$ and $B$ with position vectors $(\mathbf{3i} + \mathbf{6j} - \mathbf{8k})$ and $(\mathbf{8j} - \mathbf{6k})$ respectively, relative to a fixed origin.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for $l_1$. [2]
\end{enumerate}
The line $l_2$ has vector equation
$$\mathbf{r} = (-\mathbf{2i} + \mathbf{10j} + \mathbf{6k}) + \mu(\mathbf{7i} - \mathbf{4j} + \mathbf{6k}),$$
where $\mu$ is a scalar parameter.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that lines $l_1$ and $l_2$ intersect. [4]
\item Find the coordinates of the point where $l_1$ and $l_2$ intersect. [2]
\end{enumerate}
The point $C$ lies on $l_2$ and is such that $AC$ is perpendicular to $AB$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the position vector of $C$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [14]}}